Equilibrium and asymptotically stable equilibrium Consider the system composed by these two trajectories:
$x(t+1) = -3x(t) + \frac{7}{4}y(t)$ 
$y(t+1) = x(t)$
We have an equilibrium when $x(t) = x^0$ and $y(t) = y^0$ for each $t$. Given that I am considering an autonomous linear homogeneous system we have that $x(t+1) = Ax(t)$, where $A$ is the matrix of coefficients. An equilibrium is defined by $X^0 = AX^0$. Given that $A$ is nonsingular, the only possible equilibrium is $X^0 = (x^0,y^0) = (0,0).$ 
The equilibrium is asymptotically stable if $\lim_{t \to +\infty} x(t) = x^0$ and $\lim_{t \to +\infty} y(t) = y^0$, which in this case is false. I do not understand how is it possible that if $x(t) = x^0$ and $y(t) = y^0$ for each $t$ the limits are not satisfied. If the trajectories are constant at the equilibrium value, how is it possible that their limit as $t$ goes to $+ \infty$ is not respectively $x^0$ and $y^0$?
 A: It can diverge to infinity (in norm) or follow a periodic orbit:
There's nothing special about your matrix: consider, for example, the simpler "decoupled" case of
$$A = \left(
\begin{array}{cc}
 2 & 0 \\
 0 & 2 \\
\end{array}
\right).$$
I.e., $x(n+1) = 2x(n),\qquad y(n+1) = 2y(n)$.
You can see that in this case, every initial condition besides $(x(0),y(0)) = (0,0)$ diverges to $∞$.
(Another viewpoint is that $(x(n),y(n)) → (0,0)$ as $n → -∞$, since $A$ is invertible. You can't get both going on in this linear setting!)
$$A = \left(
\begin{array}{cc}
 0 & 1 \\
 1 & 0 \\
\end{array}
\right)$$
gives another silly example: here, since $A^2 = I_2$, points generally follow a periodic orbit (and you can have more fun with rotation matrices).
It is also not true that the non-singularity of $A$ implies that (0,0) is the only equilibrium. Consider $A = I_2$ for a simple counterexample.
In your case, because the eigenvalues of $A$ are $-7/2$ and $1/2$, i.e.,
$$A \cdot (7,-2) = - \frac 72 (7,-2). $$
One has, by linearity, that #
$$A^n (7ε, -2ε) = \left(-\frac 72\right)^n (7ε, -2ε)$$
which diverges for $ε > 0$ arbitrarily close to zero(i.e., points arbitrarily close to 0 diverge).

In general,

*

*If $A$ has 1 as an eigenvalue, $(0,0)$ won't be the only equilibrium.


*if $A$ has an eigenvalue that is greater than 1 in size, generically, orbits will diverge to infinity.
A: All homogeneous linear system have $(0,0)$ the origin of coordinates as an equilibrium point. It can be stable or unstable. In our case
$$
X_{k+1} = A X_k,\ \ A = \left(
\begin{array}{cc}
 -3 & \frac{7}{4} \\
 1 & 0 \\
\end{array}
\right),\ \ X_k = (x_k, y_k)'
$$
$A$ has eigenvalues $(-\frac 72, \frac 12)$ one being associated to an unstable mode $|-\frac 72| > 1$ and another to a stable mode. Resuming, the origin is unstable.
